Physics

Electrons and Photons

127 Questions

Electrons and photons are fundamental concepts in modern physics, focusing on quantum interactions and the photoelectric effect. The topic involves calculating work functions, kinetic energy, and photon wavelengths. These physics questions are highly relevant for competitive test preparation.

Photoelectric effectWork function calculationsPhoton energy and wavelengthElectron emissionAtomic transitions

Electrons and Photons Questions

Multiple choice physics quantum physics photons concept of photon photons and photoelectric effect

A beam of light has two wavelengths $4971\mathring{A}$ and $6216\mathring{A}$ with a total intensity of $3.6\times { 10 }^{ -3 }W{ m }^{ -2 }$ equally distributed among the two wavelengths. The beam falls normally on an area of $1{cm}^{2}$ of a clean metallic surface of work function $2.3eV$. Assume that there is no loss of light by reflection and that each capable photon ejects one electron. The number of photo electrons liberated in $2s$ approximately :

  1. $6\times { 10 }^{ 11 }$
  2. $9\times { 10 }^{ 11 }$
  3. $11\times { 10 }^{ 11 }$
  4. $15\times { 10 }^{ 11 }$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
${ E } _{ 1 }=\cfrac { 1242 }{ 497.2 } =2.50eV;{ E } _{ 2 }=\cfrac { 1242 }{ 6621. } =2.0eV$
so, photoelectron emission takes place only due to first wavelength
$\therefore$ No. of photoelectrons emitted $=\cfrac { 1.8\times { 10 }^{ -3 }\times { 10 }^{ -4 }\times 2 }{ 2.5\times 1.6\times 10^{-19} } hv=9\times { 10 }^{ 11 }$
Multiple choice physics quantum physics photons concept of photon photons and photoelectric effect

The threshold frequency for a metallic surface corresponds to an energy of $6.2eV$, and the stopping potential for a radiation incident on this surface $5V$. The incident radiation lies in.

  1. X-ray

  2. ultra-violet region

  3. infra-red region

  4. visible region
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$hv=5eV+6.2eV=11.2eV$
$\lambda =\cfrac { 1242 }{ 11.2 } nm=1109\mathring { A } $
it lies in ultraviolet region

Multiple choice physics quantum physics photons concept of photon photons and photoelectric effect

In a photoelectric cell, illuminated with a certain radiation, the minimum negative anode of potential with respect to emitting metal required to stop the electron is $2 V.$ the  minimum KE of the photoelectrons is 

  1. $0 eV$
  2. $1 eV$
  3. $2 eV$
  4. $ 4 eV$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given,

$V _0=2V$
The minimum kinetic energy of the photo electron is
$K _{min}=eV _0$
$K _{min}=2eV$
The correct option is C.

Multiple choice physics quantum physics photons concept of photon photons and photoelectric effect

Assuming photo-emission to take place, the factor by which the maximum velocity of the emitted photo electrons changes when the wavelength of the incident radiation is increased four times, is (assuming work function to be negligible in comparison to $hcl\lambda $)

  1. 4
  2. $\dfrac{1}{4}$

  3. 2
  4. $\dfrac{1}{2}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Einstein's Photoelectric equation is given as : $\dfrac{hc}{\lambda } = \phi + \dfrac{1}{2} mv^{2}$ (where $\phi $ = work function , v = velocity of photoelectron, $\lambda $ = wavelength of incident radiation) 

 as $\dfrac{hc}{\lambda } >>\phi \Rightarrow \dfrac{hc}{\lambda }\approx \dfrac{1}{2} mv^{2}$

 now wavelength is increased by 4 times : 

 $\Rightarrow \lambda _{2}=4\lambda $ 

 $\Rightarrow \dfrac{hc}{4\lambda }=\dfrac{1}{2} mv _{2}^{2}$

 $\Rightarrow \dfrac{1}{4}\left ( \dfrac{1}{2} mv^{2}\right )=\dfrac{1}{2}mv _{2}^{2}$ 

 $v _{2}^{2}=\dfrac{v^{2}}{4}\Rightarrow v _{2}=\dfrac{v}{2}$ 

 so maximum velocity of photoelectrons will be $\dfrac{1}{2}$ times when wavelength becomes 4 times.

Multiple choice physics quantum physics photons concept of photon photons and photoelectric effect

Radiational wave length $ \lambda $=124 nm falls on a metallic surface. Then the kinetic energy of the ejected photo electron(s) can be : (Given that threshold wavelength ($ \lambda _{0} $)=248 nm)

  1. 1 eV

  2. 2 eV

  3. 3 eV

  4. 5 eV
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
We know,
$KE=hv-h{ v } _{ 0 }$
        $=hc\left[ \dfrac { 1 }{ \lambda  } -\dfrac { 1 }{ { \lambda  } _{ 0 } }  \right] $
$=6.626\times { 10 }^{ -34 }\times 3\times { 10 }^{ 8 }\times \left[ \dfrac { 1 }{ 124\times { 10 }^{ -9 } } -\dfrac { 1 }{ 248\times { 10 }^{ -9 } }  \right] $
$=0.08015\times { 10 }^{ -17 }$
$=8.015\times { 10 }^{ -19 }J$
$1.6\times { 10 }^{ -19 }J=1eV$
$8.015\times { 10 }^{ -19 }J=\dfrac { 1 }{ 1.6\times { 10 }^{ -19 } } \times 8.015\times { 10 }^{ -19 }$
                            $=4.74eV$
Thus, the answer is close to $5eV$.
Multiple choice physics quantum physics photons concept of photon photons and photoelectric effect

Violet light is falling on a photosensitive material causing ejection of photoelectrons with maximum kinetic energy of $1$ eV. Red light falling on metal will cause emission of photoelectrons with maximum kinetic energy (approximately) equal to

  1. $1.2$ eV
  2. $0.9$ eV
  3. $0.5$ eV

  4. Zero, that is no photoemision
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The photoelectric equation is K_max = E_photon - Work Function. Violet light has a shorter wavelength and higher energy than red light. If red light has a frequency lower than the threshold frequency of the metal, no photoelectrons will be emitted.

Multiple choice physics quantum physics photons concept of photon photons and photoelectric effect

The work function of a metal is $3.3\times{  10}^{  -19}J$. The maximum wavelength of the photons required to eject electron from the metal is:

  1. $200nm$
  2. $300nm$
  3. $400nm$
  4. $600nm$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The maximum wavelength corresponds to the threshold wavelength where photon energy equals the work function. Using lambda = hc / W, where h = 6.626 * 10^-34 J s and c = 3 * 10^8 m/s, we get lambda = (6.626 * 10^-34 * 3 * 10^8) / (3.3 * 10^-19) = 6 * 10^-7 m = 600 nm.

Multiple choice physics quantum physics photons concept of photon photons and photoelectric effect

A metal plate of area $1\times { 10 }^{ -4 }{ m }^{ 2 }$ is  illuminated by a radiation of intensity 16 m $W/{ m }^{ 2 }.$ The work function of the metal is 5 eV. The energy of the incident photons is 10 eV and only 10% of it produces photo electrons. The number of emitted photo electrons per second and their maximum energy, respectively, will be :
$\left[ { 1eV=1.6\times 10^{ 19 }J } \right] $

  1. ${ 10 }^{ 12 } and\ 5eV$
  2. ${ 10 }^{ 11 } and\ 2.5eV$
  3. ${ 10 }^{ 10 } and\ 5eV$
  4. ${ 10 }^{ 14 } and\ 5eV$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice physics quantum physics photons concept of photon photons and photoelectric effect

When radiations of wavelength 3000 are incident on a photosensitive surface, the kinetic energy of electrons is 2.5 eV. The stopping potential for 1500 will he,

  1. $V _ { s } = 2.5 \mathrm { V }$
  2. $V _ { s } = 5.0 \mathrm { V }$
  3. $2.5 \leq \mathrm { V } _ { \mathrm { s } } \leq 5.0 \mathrm { V }$
  4. $V _ { s } > 5.0 \mathrm { V }$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Energy of photon E = hc/lambda. E1 = hc/3000, E2 = hc/1500 = 2 * E1. K_max = E - Work Function. K1 = E1 - phi = 2.5 eV. K2 = 2 * E1 - phi = 2 * E1 - (E1 - 2.5) = E1 + 2.5. Since E1 > 0, K2 > 2.5 eV. The stopping potential V_s = K_max/e, so V_s > 2.5 V.

Multiple choice physics quantum physics photons concept of photon photons and photoelectric effect

A photoelectric surface is illuminated successively by monochromatic light of wavelength $\lambda$ and $\cfrac{\lambda}{2}$. If the maximum kinetic energy of the emitted photoelectrons in the second case is $3$ times that in the first case, the work function of the surface of the material is  ($h=$ Planks constant, $c=$ speed of light)

  1. $\cfrac{hc}{3\lambda}$
  2. $\cfrac{hc}{2\lambda}$
  3. $\cfrac{hc}{\lambda}$
  4. $\cfrac{2hc}{\lambda}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

K1 = hc/lambda - phi. K2 = hc/(lambda/2) - phi = 2hc/lambda - phi. Given K2 = 3 * K1, we have 2hc/lambda - phi = 3 * (hc/lambda - phi). 2hc/lambda - phi = 3hc/lambda - 3phi. 2phi = hc/lambda. phi = hc/(2lambda).

Multiple choice physics quantum physics photons concept of photon photons and photoelectric effect

In photoelectric effect for silver threshold is $\lambda _0 = 3250 \times 10^{-10} m$. If U.V of $\lambda = 2536 \times 10^{-10}$ is incident then velocity of electron from will be 

  1. $6 \times 10^6$
  2. $3 \times 10^6$
  3. $6 \times 10^5$
  4. $3 \times 10^5$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Work function phi = hc/lambda_0. K_max = hc/lambda - hc/lambda_0. K_max = hc * (1/lambda - 1/lambda_0). Substituting values: phi = (6.6 * 10^-34 * 3 * 10^8) / (3250 * 10^-10) = 6.09 * 10^-19 J. E_incident = (6.6 * 10^-34 * 3 * 10^8) / (2536 * 10^-10) = 7.8 * 10^-19 J. K_max = 1.71 * 10^-19 J. v = sqrt(2K/m) = sqrt(2 * 1.71 * 10^-19 / 9.1 * 10^-31) = 6 * 10^5 m/s.

Multiple choice physics quantum physics photons concept of photon photons and photoelectric effect

Find the maximum $KE$ of photoelectrons emitted from the surface of lithium$(\phi=2.39 eV)$ when exposed with $\displaystyle E=E _{0}(1+\cos 6\times10^{14}t)\cos 3.6\times 10^{15}t$

  1. 0.37 $eV$
  2. 0.1 $eV$
  3. 0.02 $eV$
  4. 0.06 $eV$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For maximum $KE.$ frequency should be maximum. Here,
$\displaystyle w=3.6\times10^{15}s^{-1}$


$\displaystyle (KE) _{max}=\dfrac{hw}{2\pi}-\phi$

$=\displaystyle =\dfrac{6.625\times10^{-34}\times3.6\times10^{+15}}{6.28\times1.6\times10^{-19}}-2.39$

$=2.36-2.39$
$=0.02$ $eV$
So, the answer is option (C).

Multiple choice physics quantum physics photons concept of photon photons and photoelectric effect

Two separate monochromatic light beams A and B of the same intensity are falling normally on a unit area of a metallic surface. Their wavelengths are $\lambda _A$ and $\lambda _B$, respectively. Assuming that all the incident light is used in ejecting the photoelectrons, the ratio of the number of photoelectrons from beam A to that from B is

  1. $(\lambda _A/ \lambda _B)^2$
  2. $\lambda _A/ \lambda _B$
  3. $\lambda _B/ \lambda _A$

  4. 1
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Number of photons per second N = Power / E_photon = (Intensity * Area) / (hc/lambda) = (I * A * lambda) / hc. Since intensity and area are the same, the number of photons is proportional to lambda. Thus, N_A/N_B = lambda_A/lambda_B.

Multiple choice physics quantum physics photons concept of photon photons and photoelectric effect

When light is made incident on a surface, then photoelectrons are emitted from it. The kinetic energy of photoelectrons

  1. Depends on the wavelength of incident light

  2. Is same

  3. Is more than a certain minimum value

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The photoelectric effect related with the incident light frequency $(\nu)$ by the following equation:
$E _k=eV _s=h\nu-\phi$,
where, $\phi$ is the work function of the material and $E _k$ is the kinetic energy of the photo electron. So the photoelectrons emitted from the surface of sodium metal are of speeds from zero to a certain maximum depending on the incident photon energy. So, the kinetic energy of photoelectrons depends on the wavelength of incident light.

So, the answer is option (A).